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Zina [86]
3 years ago
5

Please helpppp meee 20 pts Identify the arrows that show the correct direction of heat transfer.

Physics
2 answers:
kupik [55]3 years ago
7 0
Heat transfers in weird ways but it is complex i have no clue what i jus wrote but thanks fir the points luv
Finger [1]3 years ago
3 0

Answer:

The arrow pointing up from the hand and the arrow pointing diagnoly from the cup

Explanation:

That is what I THINK lmk if I'm right

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42÷7

Explanation:

6 hours

since 7km is 1hour then

42km will be 6 hours

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How do you do to your question to see if its answered?
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Go in notifications, it'll show if it was answered. If it doesn't show that a person answered it, wait a while, someone might respond :)

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It will pop up like

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What other angle would give the same range at 19 degrees?
yaroslaw [1]

Answer:

I think it is acute angle.

Explanation:

Because it is an angle between 0° and 90°. Hope this answer wil help you.

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Soapy water between surfaces
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As you travel from Detroit in a certain direction, the outside temperature, T (in degrees), depends on your distance, d (in mile
Ber [7]

Answer:

a)\Delta T= 100^{\circ}C

b)\bigtriangledown T=1^{\circ}C.mile^{-1}

c)\bigtriangledown T_4=1^{\circ}C.mile^{-1}

d)\bigtriangledown T_4=1^{\circ}C.mile^{-1}

Explanation:

Given is the data of variation of temperature with respect to the distance traveled:

Temperature T as a function of distance d:

T=(d+30) ^{\circ}C...................................(1)

(a)

Total change in temperature from the start till the end of the journey:

\Delta T= T_f-T_i..............................(2)

where:

T_f= final temperature

T_i= initial temperature

∵In the start of the journey d = 0 miles & at the end of the journey d = 100 miles.

So, correspondingly we have the eq. (2) & (1) as:

\Delta T= (100+30)-(0+30)

\Delta T= 100^{\circ}C

(b)

Now, the average rate of change of the temperature, with respect to distance, from the beginning of the trip to the end of the trip be calculated as:

\bigtriangledown T=\frac{\Delta T}{\Delta d}......................(3)

where:

\Delta d= change in distance

\bigtriangledown T=change in temperature with respect to distance

putting the respective values in eq. (3)

\bigtriangledown T=\frac{100}{100}

\bigtriangledown T=1^{\circ}C.mile^{-1}

(c)

comparing the given function of the temperature with the general equation of  a straight line:

y=m.x+c

We find that we have the slope of the equation as 1 throughout the journey and therefore the rate of change in temperature with respect to distance remains constant.

\bigtriangledown T_4=1^{\circ}C.mile^{-1}

(d)

comparing the given function of the temperature with the general equation of  a straight line:

y=m.x+c

We find that we have the slope of the equation as 1 throughout the journey and therefore the rate of change in temperature with respect to distance remains constant.

\bigtriangledown T_4=1^{\circ}C.mile^{-1}

4 0
4 years ago
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